Read the passage carefully and answer the Questions.
$KMnO_4$ is prepared by the fusion of $MnO_2$ with an alkali metal hydroxide and an oxidizing agent like $KNO_3$ to give a dark- green manganate ion which disproportionate to give permanganate as follows.
$2MnO_2+ 4KOH + O_2→2KMnO_4+2H_2O$
$3KMnO_4+ 4H^+→2KMnO_4 + MnO_2 + 2H_2O$
On heating $KMnO_4$ decomposes at 513 K to give $K_2MnO_4$. Permanganate ion is tetrahedral and diamagnetic. Acidified $KMnO_4$ acts a strong oxidizing agent which oxidizes oxalic acid, ferrous ions, nitrite ion and iodides.
Oxidation states of Mn in $MnO_2, KMnO_4$ and $K_2MnO_4$ are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → +4, +7, and +6 respectively
- MnO₂: Let the oxidation state of Mn = x
- Oxygen = –2 so 2 O = –4
$x+2(−2)=0 ⟹ x−4=0 ⟹ x=+4$
- KMnO₄: Let the oxidation state of Mn = y
- K = +1 and O₄ = –8
$+1+y+4(−2)=0 ⟹ 1+y−8=0 ⟹ y=+7$
- K₂MnO₄: Let the oxidation state of Mn = z
- 2K = +2 and O₄ = –8
$2(+1)+z+4(−2)=0 ⟹ 2+z−8=0 ⟹ z=+6$
So, the oxidation states are MnO₂ = +4, KMnO₄ = +7, K₂MnO₄ = +6.